To prove this lemma, we will use the algorithm described in [1]}.
Let \(\varepsilon _{pq}=x+y\sqrt{pq}\)  for some integers \(x\)  and \(y\) . Since \(N(\varepsilon _{pq})=1\) , then \(x^2-1=y^2pq\) . Hence by the unique factorization in \(\mathbb {Z}\)  there exist \(y_1\) , \(y_2\)  in \(\mathbb {Z}\)  such that
\((1):\ \left\lbrace  \begin{array}{ll}x\pm 1=y_1^2\\x\mp 1=pqy_2^2,\end{array}\right. \quad (2):\ \left\lbrace  \begin{array}{ll}x\pm 1=py_1^2\\x\mp 1=qy_2^2,\end{array}\right.\quad (3):\ \left\lbrace  \begin{array}{ll}x\pm 1=2py_1^2\\x\mp 1=2qy_2^2,\end{array}\right. \quad \text{ or }\quad (4):\ \left\lbrace  \begin{array}{ll}x\pm 1=2y_1^2\\x\mp 1=2pqy_2^2,\end{array}\right.\) 
Note that \(y=y_1y_2\)  or \(y=2y_1y_2\) .

System \((1)\)  implies
\(1=\left(\frac{y_1^2}{p}\right)=\left(\frac{x\pm 1}{p}\right)=\left(\frac{x\mp 1\pm 2}{p}\right) =\left(\frac{2}{p}\right)=-1.\)  So this case is impossible.

System \((3)\)  implies
\(\left(\frac{2q}{p}\right)=\left(\frac{x\mp 1}{p}\right)=\left(\frac{\pm 2}{p}\right)=\left(\frac{2}{p}\right)=-1\) , which contradicts the fact that \(\left(\dfrac{p}{q}\right)=-1\) .

If system \((4)\)  holds, then \(2\varepsilon _{ pq}=2(x+y\sqrt{pq})=2(y_1+y_2\sqrt{pq})^2\) . Thus \(\varepsilon _{ pq}\)  is a square in
\(\mathbb {Q}(\sqrt{pq})\)  which is absurd.

Suppose that \(\left\lbrace  \begin{array}{ll}x+1=py_1^2\\x-1=qy_2^2,\end{array}\right.\) 
then, \(-1=\left(\frac{p}{q}\right)=\left(\frac{x+ 1}{q}\right)=\left(\frac{x-1+ 2}{q}\right)=\left(\frac{2}{q}\right)=1.\)  Which is impossible too.

From the above discussion, we infer that \(\left\lbrace  \begin{array}{ll}x-1=py_1^2\\x+1=qy_2^2.\end{array}\right.\) 
So \( 2\varepsilon _{pq} =(\ y_1\sqrt{p}+ y_2\sqrt{q})^2\) . Therefore \(2\varepsilon _{pq}\)  is a square in \(\mathbb {Q}(\sqrt{p}, \sqrt{q})\) .
Using similar techniques, one can show that \( 2\varepsilon _{ q}\)  is a square in \(\mathbb {Q}(\sqrt{p}, \sqrt{q})\) . Hence,
\(\varepsilon _{pq}\)  (resp. \(\varepsilon _{ q}\) ) is not a square in \(\mathbb {Q}(\sqrt{p}, \sqrt{q})\) , since otherwise we will get \(\sqrt{2}\in \mathbb {Q}(\sqrt{p}, \sqrt{q})\) , which is not true.
As \(\varepsilon _{p}\)  has norm \(-1\) , it follows that \(\sqrt{\varepsilon _{q}\varepsilon _{pq}}\)  is the only element of \(\lbrace \varepsilon _{p}^i\varepsilon _{q}^j\varepsilon _{pq}^k: i, j \text{ and } k\in \lbrace 0, 1\rbrace \rbrace \)  which is a square in \(\mathbb {Q}(\sqrt{p}, \sqrt{q})\) .So the first item (cf. [1]}). We similarly prove the second item.
